User:Egm4313.s12.team11.sheider/R1.5: K 2011 p.59 pb. 5

Problem: Find a general solution to the given ODE. Check your answer by substituting into the original equation.

Given:

$${y}''+ 2\pi{y}'+\pi^2y = 0\!$$

Solution:

The characteristic equation of this ODE is therefore:

$$\lambda^2 + a\lambda + b = \lambda^2 + 2\pi\lambda  + \pi^2= 0\!$$

Evaluating the discriminant:

$$ a^2 - 4b = (2\pi)^2 - 4(\pi^2) = 4\pi^2 - 4\pi^2 = 0\!$$

Therefore the equation has a real double root and a general homogenous solution of the form:

$$ y = (c_{1}+c_{2}x)e^{-ax/2}\!$$

And finally we find the general homogenous solution:

$$ y = (c_{1}+c_{2}x)e^{-\pi x}\!$$

Checking:

We found that:

$$ y = (c_{1}+c_{2}x)e^{-\pi x}\!$$

Differentiating $$ y\!$$ to obtain $${y}'\!$$ and $${y}''\!$$ respectively:

$$ {y}' = c_{2}e^{-\pi x} - \pi (c_{1}+c_{2}x)e^{-\pi x}\!$$

And,

$$ {y}'' = -\pi c_{2} e^{-\pi x}- \pi c_{2}e^{-\pi x} + \pi^2(c_{1}+c_{2}x)e^{-\pi x}\!$$

$$ {y}'' = -2\pi c_{2} e^{-\pi x} + \pi^2(c_{1}+c_{2}x)e^{-\pi x}\!$$

Substituting these equations into the original ODE yields:

$$ -2\pi c_{2} e^{-\pi x} + \pi^2(c_{1}+c_{2}x)e^{-\pi x} + 2\pi(c_{2}e^{-\pi x} - \pi (c_{1}+c_{2}x)e^{-\pi x}) + \pi^2((c_{1}+c_{2}x)e^{-\pi x}) = 0\!$$

$$ -2\pi c_{2} e^{-\pi x} + \pi^2(c_{1}+c_{2}x)e^{-\pi x} + 2\pi c_{2}e^{-\pi x} - 2\pi^2 (c_{1}+c_{2}x)e^{-\pi x} + \pi^2(c_{1}+c_{2}x)e^{-\pi x} = 0\!$$

$$ (-2\pi + 2\pi) c_{2} e^{-\pi x} + (\pi^2 - 2\pi^2 + \pi^2)(c_{1}+c_{2}x)e^{-\pi x} = 0\!$$

$$ 0 \equiv 0 \!$$

Therefore this solution is correct.